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Curl of a vector field

by andrey21
Tags: curl, field, vector
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andrey21
#1
Feb21-11, 08:25 AM
P: 466
Find the curl of the following vector field

u = yi+(x+z)j+xy^(2)k

Now using the method Ive bin taught similar to finding determinant of 3x3 matrix here is my answer



i(2yx-1) -j(y^2) +k(0)


Just looking for confirmation if this is correct or any basic errors I have made thank you.
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Char. Limit
#2
Feb21-11, 08:29 AM
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Confirmed that it is correct, then.
andrey21
#3
Feb21-11, 08:41 AM
P: 466
Thank you Char.limit just a follow up question I am asked:

Find (curl u).v

Where
v = xi+(y^(2) - 1)j+(1-x^(2))k

Is that just the dot product of the vector v and the curl established for u. Thank you

Char. Limit
#4
Feb21-11, 08:51 AM
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P: 1,941
Curl of a vector field

Yes it is. And you can just ignore the k part completely.
andrey21
#5
Feb21-11, 08:56 AM
P: 466
So would that give me:

(2x^(2) y -x) +(y^(4) - y^(2))

Also do I still need the i j k notations??
dextercioby
#6
Feb21-11, 09:03 AM
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YOu have obtained a scalar. There's no more unit vector involved.
andrey21
#7
Feb21-11, 09:06 AM
P: 466
Oh ok so my final answer would just be 2x^(2) - x +y^(4) - y^(2)

correct?


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